Answer
$\displaystyle \frac{1}{2}$
Work Step by Step
We divide both the numerator and denominator with $\sqrt{x^{2}} =|x|,$
which, when x is negative, equals $-x.$
$\displaystyle \lim_{x\rightarrow-\infty}\frac{\sqrt{x^{2}+x}}{-2x}=\lim_{x\rightarrow-\infty}\frac{\sqrt{\dfrac{x^{2}+x}{x^{2}}}}{\dfrac{-2x}{-x}}$
$=\displaystyle \lim_{x\rightarrow-\infty}\frac{\sqrt{1+\frac{1}{x}}}{2}\qquad $... the term $\displaystyle \frac{1}{x}$ approaches 0 when $ x\rightarrow-\infty$
$=\displaystyle \frac{1}{2}$
